An infimum of a double integral on the unit disk The following question comes from Arnold's Trivium of $1991$ and it is problem $68$. I do not have a solution neither can I come up with something.
Find
$$\inf  \iint \limits_{x^2+y^2 \leq 1} \left[ \left ( \frac{\partial u}{\partial x} \right)^2 + \left ( \frac{\partial u}{\partial y} \right )^2 \right ]\, {\rm d}x \, {\rm d}y$$
for $C^{\infty}$ functions $u$ that vanish at $0$ and equal $1$ on $x^2+y^2=1$.
 A: Let $f$ be any $C^\infty$ function on $[0,1]$ vanishes identically on $[0,\frac12]$ and $f(1) = 1$.
For any $\alpha \in (0,1)$, consider following function defined on the unit disk
$$u_\alpha(x,y) = f((x^2+y^2)^{\alpha/2})$$
Since $f(t)$ vanishes identically on $[0,\frac12]$, $u_\alpha(x,y)$ is $C^\infty$ on
the unit disk. It vanishes at $0$ and $= 1$ on the unit circle. This means $u_\alpha$
is a candidate to take the infimum. Notice
$$\begin{align}
\int_{x^2+y^2 \leq 1} 
 \left[ \left ( \frac{\partial u_\alpha}{\partial x} \right)^2 + \left ( \frac{\partial u_\alpha}{\partial y} \right )^2 \right]dx dy
&= 2\pi \int_0^1 r \left(\frac{df(r^\alpha)}{dr}\right)^2 dr
= 2\pi \int_0^1 r \left( \alpha r^{\alpha-1} \frac{df(r^\alpha)}{dr^\alpha}\right)^2 dr\\
&= 2\pi \alpha \int_0^1 r^\alpha \left(\frac{df(r^\alpha)}{dr^\alpha}\right)^2 dr^\alpha
= 2\pi \alpha \int_0^1 s \left(\frac{df(s)}{ds}\right)^2 ds
\end{align}
$$
We find
$$0 \le \inf_u \int_{x^2+y^2 \leq 1} 
 \left[ \left ( \frac{\partial u}{\partial x} \right)^2 + \left ( \frac{\partial u}{\partial y} \right )^2 \right]dx dy
\le \inf_{\alpha \in (0,1)} \left\{ 2\pi \alpha \int_0^1 s \left(\frac{df(s)}{ds}\right)^2 ds \right\} = 0$$
i.e. the desired infimum is $0$.
